Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I believe that I have shown that $(2,3)$ is non-principal in $\mathbb{Z}[x]$. My outline goes something like this: Assume that $(2,3) = (f(x))$ then $f(x)$ divides 2 and 3, that is 2 = $f(x)g(x)$ and 3 = $f(x)q(x)$, so the sum of the degrees of $f(x)$ and $g(x)$ is 0 and the same is true for $f(x)$ and $q(x)$. Let $f(x)=c$ where $c$ is a constant. This would then mean that $(2,3) = (\pm 1) = \mathbb{Z}[x]$, a contradiction.

I was wondering whether this argument might work for the case of $\mathbb{Q}[x]$ but it seems that you could form any polynomial from $\mathbb{Q}[x]$ using the generator $(2,3)$.

share|cite|improve this question
To the first paragraph: $3 - 2 = 1$ is contained in the ideal $(3, 2)$ of $\mathbf Z[x]$. So it seems to me that this is the unit ideal in $\mathbf Z[x]$, hence in $\mathbf Q[x]$. – Dylan Moreland Mar 30 '12 at 23:11
Indeed. Your "contradiction" is no contradiction at all: $(2,3)=(1)$, in $\mathbb{Z}[x]$ as in $\mathbb{Z}$. – Chris Eagle Mar 30 '12 at 23:12
Oh thanks for noticing that. I overlooked the obvious! – Low Scores Mar 30 '12 at 23:18
I think you overlooked the obvious because you seem to have wrote "a contradiction" like a way to end a proof. I know it will seem trivial, but when you end a proof by saying "a contradiction", it's because something must have been contradicted ; always remember to know what is. – Patrick Da Silva Mar 31 '12 at 0:54
up vote 8 down vote accepted

Note that $(2,3) = (1)$ in $\mathbb{Z}$, hence $(2,3)=(1)$ in any larger ring (that has the same unity) as well, in particular in $\mathbb{Z}[x]$ and in $\mathbb{Q}[x]$. There is no contradiction, since in fact $(2,3)$ does contain $1$: $(2,3)$ contains the difference of $3$ and $2$, after all.

On the other hand, $(2,x)$ is non-principal in $\mathbb{Z}[x]$.

More generally: if $D$ is a Unique Factorization Domain that is not a field, and $p$ is a prime of $D$, then $(p,x)$ is not principal in $D[x]$.

On the other hand, if $D$ is a field, then every ideal of $D[x]$ is principal, since $D[x]$ is a Euclidean domain.

share|cite|improve this answer

$\mathbb{Q}[x]$ is a Euclidean Domain so all ideals $\mathbb{Q}[x]$ are principal.

share|cite|improve this answer

Note $\ $ When considering the persistence of coprimality (comaximality) it is essential to consider the persistence of $1$, i.e. that the (inclusion) homomorphism preserves $1$.$\ $ If $\rm\:(r, s) = (1)\:$ in $\rm R$ then this persists in any superring $\rm S \supset R$ that has the same $1$, but it may fail if not, e.g. $\:3-2 = 1$ $\Rightarrow$ $(3,2) = (1)$ does not persist when embedded in $\mathbb Z^2$ via $\rm\:n\to (n,0)$ since in the superring it becomes $(3,0)-(2,0) = (1,0),$ but $(1,0)\ne (1,1) = 1_{\mathbb Z^2}.\:$ So comaximality does not persist in this case. Occasionally even experienced mathematicians make serious errors by overlooking this subtlety.

share|cite|improve this answer
Lol. Superrings. =) – Patrick Da Silva Mar 31 '12 at 0:52
I'm guessing it was a bit too long for a comment; I've added the caveat. – Arturo Magidin Mar 31 '12 at 2:12
@Arturo Indeed, deja vu... – Bill Dubuque Mar 31 '12 at 2:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.